Register to reply 
Efficiency of RWD vs. AWD at high speed 
Share this thread: 
#1
Mar1708, 01:14 AM

P: 17

I want to settle an ongoing debate that has plagued some car forums I have been on for years.
Scenario: A Rear wheel drive car; 1996 M3 @210whp, against an AWD 60/40 transfer case ratio 2004 STi @220ish AWHP. 0100mph race the STi wins (AWD duh) if started at 100150mph the M3 wins (RWD duh) Drag coefficient of the STi is .29, M3 is .32 (advantage STi) Final Drive gearing is near the same Basically those details above aren't really important; What I'm trying to prove to these guys is that an AWD car (with a set distribution of power [60rear/40front]) will encounter increased loss of wheel power as the speeds increase and the engine fights to put down power through all the gears and driveshafts. The RWD will suffer from the same losses, just not as much due to having less gears, less driveshafts, and less half shafts. What I'm looking for is a theory, or equation that proves this. I know for a fact that a RWD car with less whp will be faster from this 100150mph run than the AWD car with slightly more whp (how much more till the AWD car is the same speed [in time], I do not know). For the record; I do my racing on the track (the 1996 M3 being my own). This is just to prove to some of the dyno whp brainwashed guys that think a whp number is the end all be all of real world performance figures. Please don't include AWD super car examples that transfer upwards of 100% of the power to the rear wheels to take advantage of RWD superiority at speed. I already know this is why those manufactures do this and using this example to some of the simple minded folk on said forums doesn't go far :( Any help sorting these guys out will help a bunch, I already know I'm right, just lack the equations and terminology to prove such. :) Brett 


#2
Mar1708, 03:26 PM

P: 45

I'm trying to reason it out. Because you're talking about acceleration, I believe you have to take into consideration not just the horsepower available, but the torque being put on the wheels, and thus the actual force accelerating the car forward. This can also be complicated by different sizes in the front and rear tires when you consider the awd car.
For simplicity, let's consider the ideal case where both cars' engine output is (I will do this in metric for simplicity) 100kW (roughly 133 horsepower) and 100% of the power gets put to the wheels. I'm also going to assume that all driving wheels provide equal acceleration force with no slip. Each cars will also have an assumed mass of 1000kg (around 2200lb weight) I'm also going to admit that I don't know a whole lot other than the basics of how an AWD car works, so I'm going to calculate everything by the observable horsepower at each of the wheels. For simplicity's sake I'm also going to assume each wheel is equal in size and has a 1m diameter (large, but I don't think this will matter as long as it is uniform). To start the calculations, I'll assume both cars are moving at a constant speed of 50m/s, or around 112 mph. Because v=r*w (v is velocity, r is wheel radius, w is angular speed), the angular speed of the wheels are v/r, or 50/(.5)=100 rad/s . Power, in watts, is equal to P = T*w or (power = torque * angular speed). Assuming equal torque spread between all driven wheels means that there are 2 driving wheels on the RWD and 4 on the AWD. So, the total power output of all the wheels on the RWD will be T*w*2, and for AWD will be T*w*4. However, because the transmission is 100% effecient and both cars produce 100kW power, the power in both cases has to be equal to 100kW. Therefore, each wheel on the idea AWD car only puts out half the torque of the wheels on a RWD car. For numbers sake, for AWD: P=T*w*4 or 100,000 W = T * 100 rad/s * 4, or T = 250 Nm. For RWD: 100,000 W = T * 100 rad/s * 2, or T = 500 Nm. What actually accelerates the car, however, will be the force, resulting from the torque of the wheels, on the ground. Torque is force*radius, so the force from the wheels will be the torque divided by the wheel radius. Thus, for AWD: F=T/r or 250/.5 = 500 N For RWD: F=500/.5 = 1000 N The total force acting on the car will be the sum of each wheel's force on the road. In this simplified case it will simply be the individual wheel force multiplied by the number of wheels. In this example, this total for both cars is 2000 N. Both of these cars in the ideal case have the exact same force acting to accelerate them forward. Now that I've analyzed the obvious, it is important to emphasize what this all means: There is no inherent advantage for the theoretical AWD or RWD car. Even changing the power distribution between the front and rear serves no real fundamental purpose for better acceleration. You're simply multiplying and dividing by the same fixed constant and arrive to the same answer, that the force being put down ends up being the same when you assume 100% efficiency. Even shifting 100% of the power at high speed to the rear wheels on a supercar doesn't magically change anything about the physics of the situation. I just want to make sure that is clear, you probably already know this. What DOES make a difference is the aerodynamic drag, the transmission, and the additional components in the AWD vehicle. You simply cannot beat the fact that an AWD vehicle simply has more power loss between the engine and the wheels. You can't beat entropy. A supercar most likely shifts power to the rear wheels, additionally, to provide better handling and to counteract the tendency of the front end of the car to lift while accelerating. By shutting down the drivetrain between the engine and the front wheels, additionally, the supercar will not lose the power it would while in AWD mode. In addition to loss in the drivetrain, the transmission components for the AWD car are more numerous and thus will also need to be accelerated (I don't know how much this actually matters in the long run, but the engine is having to spin up additional shafts, gears, and other components. Even if the power is provided hydraulically, there will be friction loss in the hydraulic power system). All the rotational mass of these components can have the same effect as trying to run with ankle weights. Sure, you can do it, but you can't move your legs as fast because of the additional momentum. For acceleration from 0 to an arbitrary speed, the AWD car will most likely win because it has a better traction availability (less traction demand on each wheel means that the power is spread more evenly, so slippage and loss of traction is less likely). As soon as you get out of a limited traction situation, the car with the least transmission loss will win. I cannot come up with some equation that says, yes, in this situation the AWD or RWD car will win without going through some complicated assumptions and equation gymnastics. Even then, the driver of the car is probably going to mess the nice clean numbers up. Now, if you were driving the cars with electric power, the losses might make the difference between AWD and RWD insignificant, but with mechanical and hydraulic drivetrain components, I'm pretty sure RWD is going to win the high speed race, every time. 


#3
Mar1708, 04:30 PM

Sci Advisor
PF Gold
P: 2,793

You can make all manner of assumptions to prove that you're right and he's wrong. Without modelling the whole thing, or testing it, you'll never know, because there are so many variables. A simple equation doesn't do it.



#4
Mar1708, 04:57 PM

Mentor
P: 22,298

Efficiency of RWD vs. AWD at high speed
Quick question  how do you dyno an awd car?



#5
Mar1708, 06:59 PM

P: 17

#2 post was okay but too simplified because I was looking for the forces that increase in the drivetrain as the speed increases. It's much easier for the engine to turn two wheels at 150mph than it is for the engine to turn 4 wheels at 150mph even with more power. They will both go 150, but one gets there faster... that's what im out to prove. 


#6
Mar1708, 07:00 PM

P: 17




#7
Mar1708, 07:18 PM

P: 17

Here's some information I found elsewhere that explains it 10x better than I could.
"Well it's obvious, really. At lower speeds the engine power is consumed only by the AWD system. So 0100 km/h, AWD cars put in a good showing. However, at higher speeds aerodynamic drag starts to consume more and more engine power as a greater portion of the output is put to work in overcoming this aerodynamic resistance. The added resistance of an AWD car merely compounds this wasteful consumption of power." "Of course not all AWD cars employ power sapping viscous couplings on the centre differential. Many use a planetary gear arrangement or a electrohydraulic multiclutch pack to apportion power and torque fore and aft. Do I think that AWD is the ultimate form of traction control? Well from a purely mechanical and physics point of view vs two wheel drive then my answer would be yes. But it all depends on the type of AWD. If it's a permanent 50:50 or 40:60 split then yes, I'll agree. The more ondemand type systems that are predominantly FWD and only engage when slippage occurs are good in some instances but on the whole I'd rate a permanent AWD setup as being the superior form of mechanical traction control. Naturally, when you bring advanced electronics into the picture such as throttle position sensors, yaw sensors, steering angle sensors and electronically operated differentials then AWD becomes a formidable form of vehicle dynamic (and not just traction) control." What's in bold is precisely what i'm trying to prove, because it's exactly true... just need scientific proof! 


#8
Mar1708, 07:24 PM

Sci Advisor
HW Helper
P: 8,953

And in the real world the ability to put the power down through four patches of rubber on a wet/slippy road might make more difference than the extra mechanicla inefficency.
Of course to really go fast you need one wheel drive ! 


#9
Mar1708, 07:29 PM

P: 17

Grip isn't an issue. both cars weigh the same both cars have identical final drive ratios (including wheel/tire size) both cars have identical drag coefficients 


#10
Mar1708, 07:50 PM

P: 45

You know how people put light weight flywheels in racing cars so that they can rev up faster? Same idea. Every single shaft, gear, coupling, etc. that the engine has to spin is going to slow it down. You just can't get around the fact that AWD is always going to require more of these components, and thus will always have an associated disadvantage when it comes to acceleration at high speed. 


#11
Mar1708, 07:55 PM

P: 191

I'ld think of it like this: at the high speeds almost all of the horsepower is being used to overcome drag, so has the effect of reducing engine hp.
So accelerating at the upper speeds would be the equivilent of accellerating from 0 mph with say a 10 horsepower engine in each car. So now the mechanical friction losses in the 4 wheel drive car have a much larger impact in percentage use of total available power. (ie. say it took 5 hp to overcome the additional friction for four wheel drive, then this would mean it has 50% less power available to accelerate the car: which means less force at the wheel patch, Force=Mass x Acceleration) 


#12
Mar1708, 08:00 PM

P: 45

For the first part of that quote, I don't understand the reasoning. The reasoning states that because at low speed the power is consumed only by the AWD system, it will accelerate faster. The same thing could be said for RWD cars, all the power is consumed by the RWD system. The limiting factor for acceleration from a stop or from a low speed is friction. RWD cars will spin their wheels because they have to apply nearly twice the amount of force for the same engine power in order to accelerate at full throttle. Best acceleration for the cars you are talking about probably takes place at the limit of traction, which will be without slip. Most RWD cars simply cannot put all the power to the ground because of slippage. AWD cars can because they can spread the power out among all wheels. There is a counterpoint though, and that is drag racers, which are able to spin their tires so quickly that they actually have more grip, but that is another unrelated issue. Another tangent is that some AWD cars have difficulty with full power acceleration from stop because their wheels don't slip as easily and the stress of sudden torque on all four wheels can damage components. RWD cars allow the cheap tires to be the weak link, thus avoiding twisting the frame as much. I've seen some videos of people with improperly secured rear ends that completely break off the car because they have too much traction though. As for the viscous coupling remark, ANY kind of power transfer arrangement has an associated power loss and additional rotating mass to accelerate. Some may be more advantageous than others for different reasons, but AWD will always have more power loss associated with it (unless the RWD designer is completely incompetent or deliberately handicaps the RWD car). 


#13
Mar1708, 08:06 PM

P: 45

Another thing to add, to put numbers behind the friction argument:
For an AWD car: The numbers before were 500 N for each wheel. The car weighs 1000kg, and assuming an even spread of weight among all 4 wheels puts 250kg, or roughly 2500N on each tire. This means that to accelerate at full throttle without slip, the tires will need a friction coefficient of 500/2500 or .2 . For RWD: Each wheel will need 1000 N of acceleration force, and the weight per wheel will be the same as the AWD car, or 2500 N. This means the required coefficient of friction for acceleration at full throttle for the RWD car will be 1000/2500, or .4, TWICE AS MUCH! Granted, the car's weight will shift towards the back of the car and thus allow a smaller coefficient of friction for the RWD car, but unless the car is standing on it's rear wheels, it's still going to slip before the AWD car. 


#14
Mar1708, 08:12 PM

P: 191

further to my last post
using some real accerlation values, one set for low speed, one set for high speed, you could calculate how much HP is consumed by drag at speed by F=m x a, get the 'force' which is accelerating the car, then sub in to: Power (hp) = Force (lb) * Velocity (MPH) / 374 giving the available horsepower for acceleration (ie. approx. size of the 'small' engine) then you could find out how much horsepower is consumed by driving the additional 2 wheels, and then calculate how much of the available horsepower this 'wastes' 


#15
Mar1708, 08:19 PM

P: 17

Greg, I know how cars work. I've done many many many passes at the drag strip. I'm not looking for an explanation of how things work; I've already stated that. I just need a top speed calculation of power losses as the car fights against drag (internal and external) forces at speed.
Homer; post #14 was precisely what I was looking for, just need to factor in the additional 2 wheels. Can you come up with a brief equation so I can chart out power availability every 10mph increased (increment not important) for the AWD car. 


#16
Mar1708, 08:22 PM

P: 17

The extra 2 wheels means there are 2 more half shafts (axles), one more diff, and a transfer case (where the power is set at 60% rear/40% front, or 50/50 if it makes it easier). Some cars use a 2nd driveshaft, but for arguments sake will stick with the one.



#17
Mar1708, 08:35 PM

P: 45




#18
Mar1708, 08:35 PM

P: 17

Homer, maybe you didn't finish the equation or I am miss understanding.
When you chart that out it says power increases as speed increases. I'm guessing that's a formula for calculating how much power is needed to go X amount of speed. Need to know how much power is lost as speed rises due to driveline setup (rwd vs awd). Ideally the equation would also include a time, ie: With RWD, with these properties (weight, calculated internal drag forces [gears], tire size, drag CO, etc) the said car will go from 100150mph in x amount of time. With AWD, with these properties (weight, calculated internal drag forces [gears], tire size, drag CO, etc) the said car will go from 100150mph in x amount of time. 


Register to reply 
Related Discussions  
Collisions at high speed vs normal speed  General Physics  5  
Transformer efficiency at low (and high) frequencies  Electrical Engineering  4  
Looking for a super highefficiency air pump  General Engineering  20 